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How is v=u +at derived graphically??

Answers

By definition, a = (v-u)/t when acceleration is constant. This can be written as v = u + at or at + u. If we plot t on x axis and v on y axis. we get a straight line with intercept u and slope a. Now for any v-t graph the area under curve from any time t1 to t2, gives us the distance traveled, because it is infinite sum of infinitesimal rectangles of sides v and dt where dt tends to zero. If we consider the area under the curve, v = u +at from t =o and t = t, we get a trapezium with parallel sides u an u+at with distance between parallel lines = t s = area of trapezium = 0.5*(u+u+at)*t = ut + 0.5*(at^2) Substituting for t from first equation of motion we get the last equation of motion s = u*[(v-u)/a] + (1/2)*a*[{(v-u)/a}^2]; multiplying by 2a, we have 2as = 2uv - 2u^2 + v^2 + u^2 - 2uv = v^2 - u^2 or v^2 = u^2 + 2as